A uniformly charged non conducting disc with surface charge density 10 nC/m2 having radius 3 cm. Then find the magnitude of electric field at a point on the perpendicular bisector at a distance 2 cm from origin of the disc.

Question

A uniformly charged non conducting disc with surface charge density 10 nC/m2 having radius 3 cm. Then find the magnitude of electric field at a point on the perpendicular bisector at a distance 2 cm from origin of the disc. (A) 251 N/C (B) 126 N/C (C) 502 N/C (D) 72 N/C

Tardigrade Admin · Accepted Answer

E=(σ/4πε0)(2π)[1-(x/√R2+x2)] =9×109×10×10-9×6.28[1-(2/√4+9)] ∴ E=90×6.28[1-(2/√13)]=251 N/C

Q. A uniformly charged non conducting disc with surface charge density $10 n C / m^{2}$ having radius $3 c m$ . Then find the magnitude of electric field at a point on the perpendicular bisector at a distance $2 c m$ from origin of the disc.

251 N/C

126 N/C

502 N/C

72 N/C

$E = \frac{σ}{4 π ε _{0}} (2 π) [1 - \frac{x}{R ^{2} + x ^{2}}]$

$= 9 \times 1 0^{9} \times 10 \times 1 0^{- 9} \times 6.28 [1 - \frac{2}{4 + 9}]$

$∴ E = 90 \times 6.28 [1 - \frac{2}{13}] = 251 N / C$

Q. A uniformly charged non conducting disc with surface charge density 10nC/m2 having radius 3cm. Then find the magnitude of electric field at a point on the perpendicular bisector at a distance 2cm from origin of the disc.